CHE 303 (Winter 2009)
__________________
LAST NAME, FIRST
Problem set #4
1) What pressure is needed to isothermally compress ice initially at  5oC and 1 bar so that
it changes phase? At 0 ºC and 1 bar water density is 1000 kg/m3 and ice density is 917
kg/m3. Also at 0 ºC and 1 bar, h l  h s = 6010 J/mol. (654.1 bar)
2) One mole of a pure species exists in liquid-vapor equilibrium in a rigid container of
volume V = 1 L, a temperature of 300oK, and a pressure of 1 bar. The enthalpy of
vaporization and the second virial coefficient in the pressure expansion are:
hvap = 16,628 [J/mol] and B’ =  110-7 [m3/J]
Assume the enthalpy of vaporization does not change with temperature. You may neglect
the molar volume of the liquid relative to that of the gas.
(a) How many moles of vapor are there?
(b) This container is heated until the pressure reaches 21 bar and is allowed to reach
equilibrium. Both vapor and liquid phases are still present. Find the final temperature of
this system. At this pressure you need to use the viral equation of state given by
vv 


RT 1  B ' P
1

 RT   B ' 
P
P

(c) How many moles of vapor are there now?
Ans: (a)
nv 
V

v
 
0.001 m 3
 m3 
0.0249 

 mol 
n v  0.0402 mol
(b) T2  523.3 K
(c)
nv 
V

vv
 
0.001 m 3
 m3 
0.00164 

 mol 
n v  0.61 mol 
1)
Engineering and Chemical Thermodynamics by Koretsky M.D
3) Tired of studying thermo, you come up with the ideal of becoming rich by manufacturing
diamond from graphite. To do this process at 25oC requires increasing the pressure until
graphite and diamond are in equilibrium. The following data are available at 25oC:
g(25oC, 1 atm) = gdiamond  ggraphite = 2866 [J/mol], diamond = 3.51 [g/cm3], and
graphite = 2.26 [g/cm3].
Estimate the pressure at which these two forms of carbon are in equilibrium at 25oC.
Ans: P = 1,514 [MPa] = 15,143 [bar]
4) You wish to know the melting temperature of aluminum at 100 bar. You find that at
atmospheric pressure, Al melts at 933.45oK and the enthalpy of fusion is:
hfus =  10,711 [J/mol]
Heat capacity are given by
clp = 31.748 [J/moloK], csp = 20.068 + 0.0138T [J/moloK]
Take the density of solid aluminum to be 2700 [kg/m3] and liquid to be 2300 [kg/m3]. At
what temperature does Al melt at 100 bar?
Ans: T = 934.91 [K]
5. (6.61) The vapor pressure of silver (between 1234oK and 2485oK) is given by the
following expression:
ln P = 
14,260
 0.458 ln T + 12.23, where P is in torr and T in oK.
T
Estimate the enthalpy of vaporization at 1500oK. State the assumptions that you make.
 kJ 
Ans: h vap  112.8 
 mol 
6. (6.91) At a temperature of 60.6oC, benzene exerts a saturation pressure of 400 torr. At
80.1oC, its saturation pressure is 760 torr. Using only these data, estimate the enthalpy of
vaporization of benzene. Compare it to the reported value of hvap = 35 [kJ/mol].
 kJ 
Ans: h vap  32.3 
 mol 
This value is 7.7% smaller than the reported value.
7. (5.211) A well-insulated, rigid vessel is divided into two compartments by a partition. The
volume of each compartment is 0.1 m3. One compartment initially contains 400 moles of
gas A at 300oK, and the other compartment is initially evacuated. The partition is then
1)
Engineering and Chemical Thermodynamics by Koretsky M.D
removed and the gas is allowed to equilibrate. Gas A is not ideal under these
circumstances but can be described well by the following equation of state:
P=
RT
a

, where a = 42 [JKm3/mol2] and b = 3.210-5 [m3/mol].
2
v  b Tv
You may take the ideal gas heat capacity of gas A to be cv = 1.5R. Calculate the final
temperature.
Ans:
1)
T f  261.6 K
Engineering and Chemical Thermodynamics by Koretsky M.D